I was thinking along Ohm's Law as well, combined with P = U x I (Power equals Voltage times Current). Those two combined, give P = U^2 / R (Power equals Voltage squared divided by Resistance). In other words, a resistance twice as high will result in half the power output.

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Ummmm? What? I mean Watt? And yet, your 'real world' test shows that the volume output remains [noticeably] the same. In other words, best check the battery drain.

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Remember I have not been able to establish the impedance of the nüvi 2495LMT speaker, only the one from the nüvi 2595LMT (4 ohm). From a point of standardisation across nüvi's not likely, but maybe the original speaker from the nüvi 2495LMT was 8 ohm too...

As for battery drain: if using an 8 ohm speaker instead of a 4 ohm speaker would only reduce the battery drain (but also reduce the volume output). Right?

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That's not necessarily correct in regard to the volume as you've observed anecdotally. Think of the relative electrical resistance in the same way as a water pipe's diameter will effect the flow of water. A larger pipe [less resistance] allows more water to flow, but can the pump [amplifier] cope with the demands required by the smaller pipe [HIGHER resistance] is the question. Hence my comment in Post #10 in regard to the output source.
But hey Mate, if it works ok in the nuvi then it's all good .....

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